Question

If a bullet losses 1/3 of its velocity on penetrating 1cm in a wooden block,then how much it will penetrate before coming to rest?
A.5/4 cm
B.4/5 cm
C.3/4 cm
D.4/3 cm

Answer 1

Given: velocity (v1 assumed) reduced to 1/3 of initial velocity (u1 assumed) when displacement travelled is (s1)=1cm

To find: s when final velocity is 0

Solution: This question can be solved using Newton's Laws of Motion.

Let us assume initial velocity as u1 and velocity after penetrating 1cm as v1.

Given, v1= u1- (u1/3) = (2u1/3)-------------------->(1)

Now, applying the formula

v²-u²=2as  ----------------------------------------->(2)                                        

where v= final velocity

          u= initial velocity

          a= acceleration

          s= displacement

in equation (1),

v1²-u1²= (-2a)*1   (∵ we are supposed to get the answer in cm it is not               necessary to convert.)

                        (a is taken negative as it is de-acceleration in this case)

⇒ (4u1²/9) - u1² = -2a

⇒ -5u1²/9  = -2a

⇒  5u1²/9 = 2a

⇒  u1²/2a = 9/5---------------------------------------------------------->(3)

We are asked to find the displacement when final velocity v=0 or bullet comes to rest. So here initial velocity for this time period will be u= 2u1/3.

Applying formula (2) with the new values we get the equation,

0 - (4u1²/9) = -2as   (where s is the displacement we need to find)

⇒ u1²(-4/9) = -2as

⇒  u1²(4/9) = 2as

⇒ s = (u1²/2a)*(4/9)

putting value of (u1²/2a) from equation (3),

⇒ s= (9/5)*(4/9)

⇒ s= 4/5

∴ s= 4/5 cm

ANS: 4/5 cm

(Hope you understood and liked the solution. If yes, please upvote.)

Answer 2

Given:

the velocity of the bullet (v₁ assumed) decreases to 1/3 of initial velocity (u₁ assumed) when the distance traveled is (s₁)=1cm

To find:

distance traveled before the bullet comes to rest.

Solution:

By  Newton's Laws of Motion.

Let the initial velocity be u₁ and velocity after penetrating 1cm be v₁.

Given,  

               $V1=U1-\frac{U1}{3} \\\\V1=\frac{2U1}{3}$--------------------(i)

Now, using the speed formula,

$V^{2}- U^{2} =2AS$  ------------------>(ii)    

$v^{2}- u^{2} =(-2a)1$                                    

(a is negative as it is de-acceleration)

$\frac{4u1^{2} }{9}-u^{2} = -2a\\\\\frac{-5u1^{2} }{9} =-2a\\\\\frac{5u1^{2} }{9}=2a\\\\\frac{u1^{2} }{2a} =\frac{9}{5}$

distance when final velocity v=0 or bullet comes to rest.

So here initial velocity for this time period will be u=$\frac{2u1}{3}$

Applying formula (ii)

0 - $-\frac{4u1^{2} }{9}$ = -2as   (where s is the distance)

⇒ $u1^{2}- \frac{4}{9}$ = -2as ( solving equation)

⇒  u1²($\frac{4}{9}$) = 2as

⇒ s = ($\frac{u1^{2} }{2a}$)*($\frac{4}{9}$)

substituting value of $\frac{u1^{2} }{2a}$

⇒ s=$\frac{9}{5} *\frac{4}{9}$

⇒ s= $\frac{4}{5}$

HENCE THE DISTANCE TRAVELED BY THE BULLET IN THE WOODEN BLOCK BEFORE COMING TO REST IS 4/5CM.